前言
以前拖了好久,知道现在才知道怎么求矩阵求逆。果然是因为我是个大菜鸡。
正言
其实方法很简单,直接在原矩阵右边加上一个单位矩阵,然后高斯-约旦消元之后右边的那个矩阵就是逆矩阵了。
正确性怎么说呢?应该很显然吧。。。
( ext {Code})
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define mod 1000000007
#define MAXN 405
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
int n,a[MAXN][MAXN << 1];
int quick_pow (int a,int b){
int res = 1;for (;b;b >>= 1,a = 1ll * a * a % mod) if (b & 1) res = 1ll * res * a % mod;
return res;
}
signed main(){
read (n);
for (Int i = 1;i <= n;++ i){
for (Int j = 1;j <= n;++ j) read (a[i][j]);
a[i][n + i] = 1;
}
for (Int i = 1;i <= n;++ i){
int fir = 0;
for (Int j = i;j <= n;++ j) if (a[j][i]){fir = j;break;}
if (!fir) return puts ("No Solution"),0;
if (fir ^ i) swap (a[fir],a[i]);
int kk = quick_pow (a[i][i],mod - 2);
for (Int j = 1;j <= n;++ j){
if (j == i) continue;
int Inv = 1ll * a[j][i] * kk % mod;
for (Int k = i;k <= n << 1;++ k) a[j][k] = (a[j][k] + mod - 1ll * a[i][k] * Inv % mod) % mod;
}
for (Int j = 1;j <= n << 1;++ j) a[i][j] = 1ll * a[i][j] * kk % mod;
}
for (Int i = 1;i <= n;++ i){
for (Int j = 1;j <= n;++ j) write (a[i][j + n]),putchar (' ');
putchar ('
');
}
return 0;
}